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I'm going to ask this question first in classical setting and then sketch its natural geometric setting.

Let $\Gamma$ be a subgroup of $\operatorname{PSL}_2(\mathbb Z)$ (the question is mostly interesting in the case where $\Gamma$ is a thin subgroup, i.e. non-elementary and of infinite index). Let $o$ be any point in the Poincaré half-plane $\mathbb H^2$. Define the Poincaré series $$ p_o(s) = \sum_{\gamma \in \Gamma} e^{-s \cdot d(o, \gamma o)}. $$ This always converges for $s > 1$ and the smallest $s$ for which it converges is called the critical exponent of $\Gamma$.

On the other hand let $\Gamma_\infty$ be the subgroup $$ \Gamma_\infty = \Gamma \cap \left\{ \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} :\: x \in \mathbb Z \right\} $$ and let $$ q_o(s) = \sum_{\gamma \in \Gamma/\Gamma_\infty} \operatorname{Im}(\gamma o)^s $$ which also converges for $s > 1$.

I am fairly certain that the infimal abscissa of convergence for the series $q_o(s)$ is the critical exponent of $\Gamma$, and that the behaviour at the critical exponent is the same. At least for the equality of abscissae of convergence i think this can be proved directly, by elementary estimates. I am also wondering whether one could recover $q_o$ from (variants of) $p_o$ through an asymptotic process. Does anybody know if this was written up somewhere? (The few places i looked at seem to consider both kind of series independently and not attempt to relate them.)


Of course, in the question above one can replace :

  • $\mathbb H^2$ with a proper Gromov-hyperbolic space $X$;

  • $\Gamma$ with a discrete subgroup of isometries of $X$;

  • $z\mapsto \operatorname{Im}(z)$ with (the exponential of) a Busemann function at a parabolic fixed point of $\Gamma$.

I would think that everything works fine if the stabilisers in $\Gamma$ of points at infinity of $X$ are amenable (i don't know if that is always the case in these circumstances).

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  • $\begingroup$ Why are you "fairly certain?" I would be worried about the fact that the critical exponent of $\Gamma_\infty$ is $1/2$. Did you do the actual computation? $\endgroup$ Commented Jun 14 at 15:57
  • $\begingroup$ I though that the exponential distortion of the parabolic subgroups would not play a role i the computations but maybe i'm wrong (i did not carry them over). In the case of lattices everything is fine ; in general maybe the hypothesis that the exponent is >1/2 should be added (this is always the case for normal subrgoups of lattices). $\endgroup$ Commented Jun 15 at 7:57

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